C#:获取 XML 文档的所有节点 [英] C# : Getting all nodes of XML doc

问题描述

有没有一种简单的方法可以从 xml 文档中获取所有节点?我需要每个节点、子节点等来检查它们是否具有某些属性.

Is there a simple way, to get all nodes from an xml document? I need every single node, childnode and so on, to check if they have certain attributes.

或者我是否必须爬过文档，要求子节点?

Or will I have to crawl through the document, asking for childnodes?

推荐答案

在 LINQ to XML 中非常简单:

In LINQ to XML it's extremely easy:

XDocument doc = XDocument.Load("test.xml"); // Or whatever
var allElements = doc.Descendants();

所以要查找具有特定属性的所有元素，例如:

So to find all elements with a particular attribute, for example:

var matchingElements = doc.Descendants()
                          .Where(x => x.Attribute("foo") != null);

假设您想要所有元素.如果您想要所有节点(包括文本节点等，但不包括作为单独节点的属性)，您可以使用DescendantNodes().

That's assuming you wanted all elements. If you want all nodes (including text nodes etc, but not including attributes as separate nodes) you'd use DescendantNodes() instead.

LINQ to XML 中的命名空间很好.你会使用:

Namespaces in LINQ to XML are nice. You'd use:

var matchingElements = doc.Descendants()
                          .Where(x => x.Attribute(XNamespace.Xmlns + "aml") != null);

或用于不同的命名空间:

or for a different namespace:

XNamespace ns = "http://some.namespace.uri";

var matchingElements = doc.Descendants()
                          .Where(x => x.Attribute(ns + "foo") != null);

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